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Bond [772]
3 years ago
11

Why does the Sun appear so big, bright, and hot if it is only an average sized star?

Physics
2 answers:
Oliga [24]3 years ago
4 0

Answer:

However, compared to other stars, our Sun is only a medium-sized star, meaning that some stars are much larger than the Sun and some are much smaller. The Sun looks bigger than other stars because it is so much closer to the Earth. The further away an object is, the smaller it appears, even if it is very big.

Elena L [17]3 years ago
3 0

Answer:

The Sun looks bigger than other stars because it is so much closer to the Earth. The further away an object is, the smaller it appears, even if it is very big.

Explanation:

However, compared to other stars, our Sun is only a medium-sized star, meaning that some stars are much larger than the Sun and some are much smaller.

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How much work is done to move 2.10 μC of charge from the negative terminal to the positive terminal of a 3.50 V battery?
rjkz [21]

Answer:

6.3E-6

Explanation:

Workdone = V Q

4 0
3 years ago
If a girl is standing still and holding a box, is she doing any work? Why or why not?
sukhopar [10]
I beleive she isnt doing any work due to holding the box motionless, you must be exerting a force in the direction of the box motion. If she is just standing there holding the box their isn't no work becuase no distance has been covered. work = force = distance.


7 0
3 years ago
Read 2 more answers
A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio
natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}

For the maximum velocity of the ball at the top of the vertical circular motion,

v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}

where g is the acceleration due to gravity,

Substituting the known values,

\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

8 0
1 year ago
A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth to the moon, a distance of 384,000 km. Sup
Tema [17]

Answer:

a) v = 19,149.6 m/s

b) f = 95%

c) t = 346.5min

Explanation:

First put all values in metric units:

15.8 min*\frac{60s}{1min}=948s

The equation of motion you need is:

v_f = a*t+v_0

where v_f is the final velocity, a is acceleration and t is time in hours.

Since the spaceship starts from 0 velocity:

v_f = a*t = 20.2*948 = 19,149.6 m/s

Next, you need to calculate the distances traveled on each interval, considering that both starting and final intervals travel the same distance because the acceleration and time are equal. For this part you need the next motion equation:

x=\frac{v_0+v_f}{2}t

solving for first and last interval:

Since the spaceship starts and finish with 0 velocity:

x=\frac{v}{2}t=\frac{19,149.6}{2}948=9,076,910.4m=9,076.9104km

Then the ship traveled 384,000-9,076.9104*2 = 361,846.1792km at constant speed, which means that it traveled:

f_{constant_speed} =\frac{ x_{constant_speed}}{x_total} =\frac{361,846.1792}{380,000} =0.95

Which in percentage is 95% of the trip.

to calculate total time you need to calculate the time used during constant speed:

t = \frac{361,846,179.2}{19,149.6} = 18,895.75s = 314min

That added to the other interval times:

t_{total} = t_1+t_2+t_3=15.8+314.93+15.8=346.5min

5 0
3 years ago
(a) Two ions with masses of 4.39×10^−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic fiel
sammy [17]

Answer:

Part a)

R_1 = 0.072 m

Part b)

R_2 = 0.036 m

Part c)

d = 0.072 m

Explanation:

Part a)

As we know that the radius of the charge particle in constant magnetic field is given by

R = \frac{mv}{qB}

now for single ionized we have

R_1 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(1.6 \times 10^{-19})(0.301)}

R_1 = 0.072 m

Part b)

Similarly for doubly ionized ion we will have the same equation

R = \frac{mv}{qB}

R_2 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(3.2 \times 10^{-19})(0.301)}

R_2 = 0.036 m

Part c)

The distance between the two particles are half of the loop will be given as

d = 2(R_1 - R_2)

d = 2(0.072 - 0.036)

d = 0.072 m

6 0
3 years ago
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